uva 624 CD （01背包）

  CD 

You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape space and have as short unused space as possible.

Assumptions:

• number of tracks on the CD. does not exceed 20
• no track is longer than N minutes
• tracks do not repeat
• length of each track is expressed as an integer number
• N is also integer

Program should find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD

Input 

Any number of lines. Each one contains value N, (after space) number of tracks and durations of the tracks. For example from first line in sample data: N=5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

Output 

Set of tracks (and durations) which are the correct solutions and string ``sum:" and sum of duration times.

Sample Input 

5 3 1 3 4
10 4 9 8 4 2
20 4 10 5 7 4
90 8 10 23 1 2 3 4 5 7
45 8 4 10 44 43 12 9 8 2

Sample Output 

1 4 sum:5
8 2 sum:10
10 5 4 sum:19
10 23 1 2 3 4 5 7 sum:55
4 10 12 9 8 2 sum:45


简单的01背包，不过要输出路径，第一次写感觉特别麻烦，建立二维数组记录是否选中此物体。

题意：第一个数是路程，第二个数是歌曲的个数，后面的数是听每首歌的时间。要求在这段时间上尽可能地长时间听歌，每首歌必须听完整，总共最大能听歌的时间长度。
输出所有听的歌曲的时间，顺序为输入的顺序，以及最后总共的时间。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int maxs(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int i,j,n,m;
    int a[25],dp[10005];
    int s[25][10005];
    while(~scanf("%d %d",&m,&n))
    {
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        for(i=n-1; i>=0; i--)   //反序输入，为了后面能用正序
            for(j=m; j>=a[i]; j--)   //最简单的01背包
                if(dp[j]<dp[j-a[i]]+a[i])
                {
                    // dp[j]=maxs(dp[j],dp[j-a[i]]+a[i]);  //这样写会报错！！！我也不知道为什么= = 可能是电脑的问题，纠结了一下午
                    dp[j]=dp[j-a[i]]+a[i];
                    s[i][j] =1;
                }
        for(i=0,j=dp[m]; i<n,j>0; i++)  //输出路径
        {
            if(s[i][j])
            {
                printf("%d ",a[i]);
                j-=a[i];
            }
        }
        printf("sum:%d
",dp[m]);
    }
    return 0;
}